The invention pertains to the field of telecommunication systems. More particularly, the invention pertains to the detection of false positives in frame synchronization of a Code Division Multiple Access (CDMA) telecommunications system.
Code Division Multiple Access (CDMA) is currently being used by the majority of communication carriers in the United States to provide improved digital wireless telephone service to their customers.
CDMA is the term applied to a family of digital communication techniques called xe2x80x9cspread spectrumxe2x80x9d that have been used in military applications for many years. Spread spectrum uses noise-like carrier waves and wider bandwidths than those required for simple point-to-point communication at the same data rate. The two original driving motivations to develop spread spectrum technology were to prevent jamming of communications (anti-jam, or AJ), and to hide the fact that communication was taking place, often called low probability of intercept (LPI). By spreading the communications signals over a wider bandwidth, the signal is both decreased to an extent that it is not detected by unfriendly receivers, and spread out enough so that the signal can not be jammed.
CDMA changes a basic device for a system from a predominately analog device to a predominately digital device. CDMA receivers do not eliminate analog processing entirely, but they separate communication channels by means of a modulation that is applied and removed digitally, not on the basis of frequency. Multiple users occupy the same frequency band. This universal frequency reuse is essential to the very high spectral efficiency that is the cornerstone of CDMA.
CDMA allows the transmission of multiple messages through a common channel. In multiple access, many users utilize the same communications path. All users occupy the same channel bandwidth and transmit simultaneously. But, each user has a unique wide-bandwidth code, or spreading sequence. The properties of this unique spreading sequence allows each receiver, when correctly synchronized to the spreading sequence of the user, to receive the signal of that user and block out the signals of the other users sufficiently so that interference is at a tolerable level.
In the transmission of digital data between two units in a telecommunication system, bits of data are commonly grouped in bytes, bytes are grouped in segments, segments are grouped in blocks called frames and frames are transmitted in a window of data. The beginning and end of each frame needs to be clearly identifiable. For reliable communication, the unit receiving the data must be synchronized with the unit transmitting the data.
Synchronization occurs at different levels. Bit synchronization occurs when the receiving unit correctly recognizes the boundaries between bits of received data. Frame synchronization occurs when the receiving unit correctly recognizes the frame boundaries in the received data. Synchronization similarly occurs for bytes and for segments.
Within a CDMA base station air interface, frame synchronization detection and validation is an essential function. Frame synchronization is used to detect an initial channel connection as well as a continuous run-time validation of channel connectivity. Its current status is transferred periodically along with channel receive data packets. The active synchronization condition is used to enable closed-loop power control, to enable a reduced searcher window (for shorter search cycles) and to assist in identifying valid data flow.
There are two different types of frame synchronization. First, the xe2x80x9cInternal Frame Synchronizationxe2x80x9d (IFS). This synchronization status is determined solely from checks against the known received pilot block symbol pattern. Active IFS status is used to enable closed-loop power control. The second type of frame synchronization is referred to as xe2x80x9cdata validity frame synchronizationxe2x80x9d (DVFS). This type synchronization status is used to indicate to the system whether or not the receive signal appears to have been lost. That is, inactive DVFS indicates that the receiver is no longer tracking the signal. The general algorithm to determine DVFS status uses a combination of the IFS status and the CRC (cyclic redundancy code) check result on each receive data packet.
Description of the IFS algorithm depends on understanding the received pilot block symbol pattern within the receive data frame structure. The typical data frame structure is a 10 mS packet consisting of usually 16 time slots. Within each time slot, several symbols are devoted to the pilot block sub-channel. The pilot block consists of a known sequence of symbols. The pilot symbol sequence varies depending on the time slot number, but the sequence is identical for each frame
Note that the example pattern of Table 1 is for four pilot symbols per time slot and QPSK modulation so that there are two bits per symbol. Columns (Symbol #) 1 and 3 are referred to as the synchronization word (SW) symbols. The even numbered columns are framing symbols. Note that these are typically all ones. The pilot sequence is always known to baseband processing equipment. Channel timing, determined as a normal part of CDMA receivers provides continuous knowledge of the current received time slot and symbol numbers.
The common IFS algorithm compares the received SW bits to the expected value. Parameters Nb and SR are defined for the algorithm. Internal Frame Synchronization is determined to be achieved (active) when SR or more consecutive frames are received containing Nb or fewer SW bits unmatched. The SR and Nb parameters are usually determined through simulation or experimentation. Default values are 2 frames and 10 bits, respectively. In this example, the bit compare test for every frame is a check that 10 or fewer bits mismatch out of a possible 64 bits (2 bits/symbolxc3x972 SW symbols/time slotxc3x9716 time slots per frame).
Continuing with this example, making several reasonable assumptions, the probability of a false positive indication (frame synchronization indicated when not actually true) is easily calculated. Assume that there is no valid signal present at the receiver; however, there is random noise. When noise is present with sufficient power, or as random fluctuations provide, the chip correlator complex multiply and summation may result in noisy symbol data which is roughly Gaussian with zero mean. However, each symbol is statistically independent.
The probability of a false positive result is then the cumulative probability of a binomial distribution achieving 10 hits in 64 trials:   Pf  =                    2                  -          64                    ⁢                        ∑                      k            =            0                    10                ⁢                  xe2x80x83                ⁢                  (                                                    64                                                                    k                                              )                      ≈          10              -        8            
The result is a false alarm probability of roughly 10xe2x88x928. If this number of false positives is truly correct, it should not create a problem.
However, a more thorough analysis of channel processing does indicate a problem with the frame synchronization calculation. Normal channel processing to produce valid symbol data consists of the following basic steps.
First, signal chip correlation is received to give symbols, channel estimation, symbol data channel correction and rake combining. Chip correlation de-spreads the received signal. Channel estimation uses a known symbol sequence within the pilot block in order to determine phase and magnitude changes the channel imparts. Rake combining enhances the receiver by making the symbol output a weighted average of up to eight multipath components according to each individual multipath channel estimate. These calculations, for a single rake finger only, may be written using the field of complex real numbers to represent the in-phase (I) and quadrature phase (Q) signal components of each symbol as real and imaginary parts, respectively.                     S        =                  A          ⁢                      xe2x80x83                    ⁢                      ⅇ                          j              ⁢                              xe2x80x83                            ⁢              a                                ⁢                      xe2x80x83                    ⁢          Transmitted          ⁢                      xe2x80x83                    ⁢          pilot          ⁢                      xe2x80x83                    ⁢          signal          ⁢                      xe2x80x83                    ⁢                      model            .                                              (        1        )                                H        =                  B          ⁢                      xe2x80x83                    ⁢                      ⅇ                          j              ⁢                              xe2x80x83                            ⁢              b                                ⁢                      xe2x80x83                    ⁢          Channel          ⁢                      xe2x80x83                    ⁢          transfer          ⁢                      xe2x80x83                    ⁢          function          ⁢                      xe2x80x83                    ⁢                      model            .                                              (        2        )                                Y        =                  AB          ⁢                      xe2x80x83                    ⁢                      ⅇ                          j              ⁢                              xe2x80x83                            ⁢                              (                                  a                  +                  b                                )                                              ⁢                      xe2x80x83                    ⁢          Received          ⁢                      xe2x80x83                    ⁢                      signal            .                                              (        3        )                                P        =                              ⅇ                          j              ⁢                              xe2x80x83                            ⁢              a                                ⁢                      xe2x80x83                    ⁢          Pilot          ⁢                      xe2x80x83                    ⁢          block          ⁢                      xe2x80x83                    ⁢          signal          ⁢                      xe2x80x83                    ⁢                      model            .                                              (        4        )                                M        =                              YP            *                    =                      AB            ⁢                          xe2x80x83                        ⁢                          ⅇ                              j                ⁢                                  xe2x80x83                                ⁢                b                                      ⁢                          xe2x80x83                        ⁢            Channel            ⁢                          xe2x80x83                        ⁢                          estimate              .                                                          (        5        )                                D        =                              YM            *                    =                                                    (                AB                )                            2                        ⁢                          ⅇ                              j                ⁢                                  xe2x80x83                                ⁢                a                                      ⁢                          xe2x80x83                        ⁢            Single            ⁢                          xe2x80x83                        ⁢            rake            ⁢                          xe2x80x83                        ⁢            finger            ⁢                          xe2x80x83                        ⁢            resulted            ⁢                          xe2x80x83                        ⁢            symbol            ⁢                          xe2x80x83                        ⁢                          data              .                                                          (        6        )            
These calculations are detailed for the symbols of a single pilot block of one time slot. Let the received signal symbol (I, Q) components be written as Y=(A0, A1), (S0, S1), (A2, A3), (S2, S3) with Ax indicating the (xe2x88x921, xe2x88x921) pilot block symbols and the Sx representing the received SW bits. In the same manner, let the known pilot symbol pattern be indicated by P=(P0, P1), (P2, P3), (P4, P5), (P6, P7). Note that P0, P1, P4 and P5 are always xe2x88x921.
Next, the channel estimate is calculated with complex multiplication. Recall that (I+jQ)(I+jq)*=(Ii+Qq)+j(Qixe2x88x92Iq).                               M          =                      xe2x80x83                    ⁢                                    (                                                -                                      A                    0                                                  -                                  A                  1                                            )                        +                          j              ⁢                              xe2x80x83                            ⁢                              (                                                      -                                          A                      1                                                        +                                      A                    0                                                  )                                      +                                                                    xe2x80x83                    ⁢                                    (                                                                    S                    0                                    ⁢                                      P                    2                                                  +                                                      S                    1                                    ⁢                                      P                    3                                                              )                        +                          j              ⁢                              xe2x80x83                            ⁢                              (                                                                            S                      1                                        ⁢                                          P                      2                                                        -                                                            S                      0                                        ⁢                                          P                      3                                                                      )                                      +                                                                    xe2x80x83                    ⁢                                    (                                                -                                      A                    2                                                  -                                  A                  3                                            )                        +                          j              ⁢                              xe2x80x83                            ⁢                              (                                                      -                                          A                      3                                                        +                                      A                    2                                                  )                                      +                                                                    xe2x80x83                    ⁢                                    (                                                                    S                    2                                    ⁢                                      P                    6                                                  +                                                      S                    3                                    ⁢                                      P                    7                                                              )                        +                          j              ⁢                              xe2x80x83                            ⁢                              (                                                                            S                      3                                        ⁢                                          P                      6                                                        -                                                            S                      2                                        ⁢                                          P                      7                                                                      )                                                                                      M          =                      xe2x80x83                    ⁢                                    {                                                -                                      (                                                                  A                        0                                            +                                              A                        1                                            +                                              A                        2                                            +                                              A                        3                                                              )                                                  +                                                      S                    0                                    ⁢                                      P                    2                                                  +                                                      S                    1                                    ⁢                                      P                    3                                                  +                                                      S                    2                                    ⁢                                      P                    6                                                  +                                                      S                    3                                    ⁢                                      P                    7                                                              }                        +                                                                    xe2x80x83                    ⁢                      j            ⁢                          xe2x80x83                        ⁢                          {                                                (                                                            A                      0                                        -                                          A                      1                                        +                                          A                      2                                        -                                          A                      3                                                        )                                +                                                      -                                          S                      0                                                        ⁢                                      P                    2                                                  +                                                      S                    1                                    ⁢                                      P                    2                                                  -                                                      S                    2                                    ⁢                                      P                    7                                                  +                                                      S                    3                                    ⁢                                      P                    6                                                              }                                          
Only the channel estimate symbol correction for SW bits S0 and S1 are considered here for the channel correction step.       D    =          xe2x80x83        ⁢                  (                              S            0            xe2x80x2                    +                      j            ⁢                          xe2x80x83                        ⁢                          S              1              xe2x80x2                                      )            =                        (                                                    S                0                            ⁢                              M                i                                      +                                          S                1                            ⁢                              M                q                                              )                +                  j          ⁢                      xe2x80x83                    ⁢                      (                                                            S                  1                                ⁢                                  M                  i                                            -                                                S                  0                                ⁢                                  M                  q                                                      )                                                            D          =                      xe2x80x83                    ⁢                      {                                          -                                                      S                    0                                    ⁡                                      (                                                                  A                        0                                            +                                              A                        1                                            +                                              A                        2                                            +                                              A                        3                                                              )                                                              +                                                S                  0                  2                                ⁢                                  P                  2                                            +                                                S                  0                                ⁢                                  S                  1                                ⁢                                  P                  3                                            +                                                S                  0                                ⁢                                  S                  2                                ⁢                                  P                  6                                            +                                                S                  0                                ⁢                                  S                  3                                ⁢                                  P                  7                                            +                                                                                              xe2x80x83                        ⁢                                                                                S                    1                                    ⁡                                      (                                                                  A                        0                                            -                                              A                        1                                            +                                              A                        2                                            -                                              A                        3                                                              )                                                  -                                                      S                    0                    2                                    ⁢                                      P                    3                                                  +                                                      S                    0                                    ⁢                                      S                    1                                    ⁢                                      P                    2                                                  -                                                      S                    0                                    ⁢                                      S                    2                                    ⁢                                      P                    7                                                  +                                                      S                    0                                    ⁢                                      S                    3                                    ⁢                                      P                    6                                                              }                                +                                                          xe2x80x83                    ⁢                      j            ⁢                          xe2x80x83                        ⁢                          {                                                -                                                            S                      1                                        ⁡                                          (                                                                        A                          0                                                +                                                  A                          1                                                +                                                  A                          2                                                +                                                  A                          3                                                                    )                                                                      +                                                      S                    0                                    ⁢                                      P                    2                                                  +                                                      S                    1                    2                                    ⁢                                      P                    3                                                  +                                                      S                    1                                    ⁢                                      S                    2                                    ⁢                                      P                    6                                                  +                                                      S                    1                                    ⁢                                      S                    3                                    ⁢                                      P                    7                                                  -                                                                                              xe2x80x83                    ⁢                                                    S                0                            ⁡                              (                                                      A                    0                                    -                                      A                    1                                    +                                      A                    2                                    -                                      A                    3                                                  )                                      +                                          S                0                            ⁢                              S                1                            ⁢                              P                2                                      -                                          S                1                2                            ⁢                              P                2                                      +                                          S                1                            ⁢                              S                2                            ⁢                              P                7                                      -                                          S                1                            ⁢                              S                3                            ⁢                              P                6                            }                                          
The calculations are continued in order to show the expected outcome of normal channel processing performed on random received data. For this purpose, it is desired to find the statistical expected value of the processed SW symbol (S0xe2x80x2, S1xe2x80x2). Since the Px elements are known, these values are merely constants +1 or xe2x88x921. The received SW bits are assumed to be random and statistically independent. This means that they are also uncorrelated so that E{Sa, Sb}=0 for all a and b where a is not equal to b.                               E          ⁢                      {            D            }                          =                  xe2x80x83                ⁢                              E            ⁢                          {                                                S                  0                  xe2x80x2                                +                                  j                  ⁢                                      xe2x80x83                                    ⁢                                      S                    1                    xe2x80x2                                                              }                                =                                                    P                2                            ⁢              E              ⁢                              {                                  S                  0                  2                                }                                      -                                          P                3                            ⁢              E              ⁢                              {                                  S                  0                  2                                }                                      +                          j              ⁡                              [                                                                            P                      3                                        ⁢                    E                    ⁢                                          {                                              S                        1                        2                                            }                                                        -                                                            P                      2                                        ⁢                    E                    ⁢                                          {                                              S                        1                        2                                            }                                                                      ]                                                                            =                  xe2x80x83                ⁢                                            (                                                P                  2                                -                                  P                  3                                            )                        ⁡                          [                                                E                  ⁢                                      {                                          S                      0                      2                                        }                                                  -                                  j                  ⁢                                      xe2x80x83                                    ⁢                  E                  ⁢                                      {                                          S                      1                      2                                        }                                                              ]                                ⁢                      xe2x80x83                    ⁢          BIASED                    
For the typical pilot block sequence, 18 out of 32 SW symbols have P2 not equal P3. The E{Sx2} are always greater than or equal to zero, and have significant values whenever noise power appears out of the correlator. Notice that this result does not show the desired zero expected value.
Next, the impact of the unbiased SW bit processing is determined. The important consideration for frame synchronization detection is the probability of a false positive indication from the detector; that is, the probability that frame synchronization is indicated to be true when noise is received. Since magnitude is not a consideration for the detector, the algorithm simply compares the sign of the calculated Sx with the expected Px. Mismatched signs are counted towards the Nb or fewer bits (per frame) synchronization requirement. PFx represents the probability of a false match occurring on SW bit Sx.
For S0,
PF0=Prob{sgn(P2)=sgn(S0xe2x80x2)}=Prob{sgn(P2)=sgn(P2xe2x88x92P3)|(P2!=P3)}=1
Similarly for S1,
PF1=Prob{sgn(P3)=sgn(S1xe2x80x2)}=Prob{Sgn(P3)=sgn(P3xe2x88x92P2)|(P2!=P3)}=1
This calculation is performed for all SW bits. These results indicate that the statistically biased processing of the detector is always in the direction of a false positive outcome. The bias becomes significant for SW symbols in quadrants 2 and 4 (which are the slight majority), while correlator output has greater than zero variance.
Test results show that the frame synchronization detector frequently gives a false positive result even when small levels of noise are present. The rate of this occurrence is far greater than the presumed 10xe2x88x928 rate.
These false IFS active indications very significantly disturb system performance. This is especially true for channel operation such as packet mode where rapid and verifiable data flow stop/start is critical to efficient throughput. Also, for channel types in which CRC coverage is lower. For example, an 8 bit CRC means that for all operation with false IFS indication, one out of 256 frames will be indicated good data. This places a strain on higher layer system software to deal with random data marked good. False IFS indications will also activate high-speed closed-loop power control with erroneous power control commands. This adds significantly extra noise to each system sector, which reduces system capacity.
It is apparent that channel processing and frame synchronization, as currently implemented, are flawed in terms of frequently producing a false positive. Aside from the detailed analysis above, a simple argument can be made for the problem. A channel detector is used which first assumes a known pattern is present on the channel in order to determine a channel estimate which is then used to correct the channel data. The processed data is then tested for the same known pattern. This is a circular reasoning algorithm which predetermines a positive outcome. A solution is needed to decrease the number of false positives present in the current method of frame synchronization.
Briefly stated, a solution to a problem of false positives in frame synchronization in telecommunication systems involves generating a separate channel estimate based only on the non-SW (all xe2x88x921) bits of a pilot sequence. This non-SW channel estimate is used to correct received pilot SW symbols. These specially corrected symbols are rake combined separately from other symbols. The result is used exclusively for the frame synchronization detector. Normal channel estimate calculations and rake combining proceeds as usual. This method of frame synchronization only refers to SW bit matches. False positives are decreased using the invention.